Problem Nine (Jan 19th): Let with standard orthorgonal basis , and define to be the set of all vectors . Let W be the group generated by the reflections in sending to its negative while fixing pointwise the hyperplane that is orthorgonal with . Prove that W is the semidirect product of the symmetric group and .

Solutions to last problem: be a prime, and let denote the congruence classes in modulo . also admits multiplication, making a finite field.

Now we fix a and define . Then is a subgroup of , and we can split as coset space .

When , color x with . This gives an coloring of . If , then by theorem 2 there is a monochromatic Schur triple, that is, there are integers , none of which is divisible by p, such that , since is not divisible by it follows that , completing the proof.

Theorem 1. Let . Then there exists an integer such that for all primes the congruence has a solution in the integers, such that does not divide .

Theorem 2. (Schur’s theorem)Let . Then there is a natural number , such as if and if the numbers are colored with colors, then there are three of them of the same color satisfying the equation: .

Solutions to last problem: Suppose a abstract simplicial have vertices . We say are geometric independent when the vectors are linearly independent. We first show the fact that we can find in points such that any of them are geometric independent.

We induct on the number and suppose that we have already found points such that any (or less) vertices are geometric independent. Since any (or less) vertices are geometric independent, they have to lie in a space of at most dimension. These dimensional subspaces cannot make full of the whole . We then choose a point which doesn’t lie in any of these dimensional subspaces.Thus we complete our induction by choosing an additional making any (or less) vertices choosing from are geometric independentm, completing the induction step.

Then we can choose points to corespond the abstract vertices . We now verify they are the geomeric realization of the abstract simplicial . We only need to check the the intersection of any two simplices is a face of both and . This is easily seen because there are totally no more than points involved and thus can form a simplicial . This means is either empty set or a common face of and .

Solutions to last problem: Denote the length of a permutation as . On one hand, every transposition can change the number of invrsions by one. So one easily sees that the length of a permutation is no less than the number of “inversions” in . Read the rest of this post »

George Andrews gives a new definition, namely, “concave compositins”. There are several different kinds of concave compositions. He then derived the generating functions of all these concave compositions. I wrote a post about this months ago.

Concave compositions of even length is the ordered pair of partitions with distinct parts with the property that they have the same length and that the last part are equal. We also note that zero part is allowed, too.

Recently, I figured out a combinatorial proof of the generating function of concave compositions of even length. I just submitted this work. I include the first draft of this paper here.

(Due to my business laziness, this week is a little long, longer than a month.)

Problem 6: Denote the transpositions in the symmetric group as . Define the length of a permutation as the least number of transpositions that formulate the permutation: . Show that the length of a permutation is the number of “inversions”: the number of pairs $latx i<j$ for which .